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A constant force F = m(2)g//2 is applie...

A constant force `F = m_(2)g//2` is applied on the block of mass nij as shown. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of `m_(1)`

A

`(m_(2)g)/(2m_(1))`

B

`(m_(2)g)/(2(m_(1) +m_(2))`

C

`(3m_(2)g)/(2(m_(1) + m_(2))`

D

`(3m_(2)g)/((2m_(1))`

Text Solution

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The correct Answer is:
B
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Knowledge Check

  • A constant force F = m_(2)g//2 is applied on the block of mass m_(1) as shown. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m_(1) .

    A
    `(m_(2)g)/(2m_(1))`
    B
    `(m_(2)g)/(2(m_(1)+m_(2)))`
    C
    `(3 m_(2)g)/(2(m_(1)+m_(2)))`
    D
    `(3m_(2)g)/(2m_(1))`
  • A horizontal force F is applied to a block of mass m on a smooth fixed inclined plane of inclination theta to the horizontal as shown in the figure. Resultant force on the block up the plane is:

    A
    `F sin theta + mg cos theta`
    B
    `F sin theta - mg cos theta`
    C
    `F sin theta + mg cos theta`
    D
    `F cos theta - mg sin theta`
  • In the figure shown, the blocks have equal masses. Friction, mass of the string and the mass of the pulley are negligible. The magnitude of the acceleration of the centre of mass of the blocks is (Acceleration due to gravity = g).

    A
    `((sqrt(3)-1)/(sqrt(2)))g`
    B
    `(g)/(2)`
    C
    `(sqrt(3)-1)g`
    D
    `((sqrt(3)-1)/(4sqrt(2)))g`
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